Question

PLEASE HELP!! I REALLY NEED HELP

Answer 1

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

56-57: NaCl

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   $\frac{22.99}{58.443}$ = .393

2(Cl): $\frac{35.453}{58.443}$= .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

58-60 $K_{2} CO_{3}$

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: $\frac{78.196}{138.204}$= .566 Step 3: (.566)(100)= 56.6%

2: C: $\frac{12.011}{138.204}$= .087 Step 3: (.087)(100)= 8.7%

2: O: $\frac{47.997}{138.204}$= .347 Step 3: (.347)(100) = 34.7%

61-62 $Fe_{3} O_{4}$

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: $\frac{167.535}{231.531}$= .724 Step 3: (.724)(100)= 72.4%

2: O : $\frac{63.996}{231.531}$= .276 Step 3: (.276)(100) = 27.6%

63-65 $C_{3}H_{5}(OH)_{3}$

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: $\frac{36.033}{92.094}$= .391 Step 3: (.391)(100) = 39.1%

2: H: $\frac{8.064}{92.094}$= .088 step 3: (.088)(100) = 8.8%

2: O: $\frac{47.997}{92.094}$ = .521 step 3: (.521)(100) = 52.1%